HW#5%20solution%20corrected

# HW#5%20solution%20corrected - V 15 m/s.— Area = 0.3 m2...

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Unformatted text preview: V: 15 m/s .— Area = 0.3 m2 r \ ,.. \‘ Area=1m2 CV .42" if}; Water ﬂows as two free jets from the tee attached to the pipe shown in Fig. P533. The exit speed is 15 m/s. If viscous effects and gravity are negligible, determine the x and M” y components of the force that the pipe exerts on the tee. . _ , Area z 0.5 m2 M56 ‘H’Ié COHﬁ/a/ volume, Shown. Far 'H’Ie X—cam’aonem‘ of #6: 199m: ﬁxer/{0’ by +116 Pllpé’. 0n 'f/ve +86. we use, #18. x— wmpanenf of #26 Izhear manden7‘um egaaﬁmn. “MFVAI4’VW0K’A3 :KA,'gA3—%5{Ad’qz)+f ‘féap‘zwﬂr/g; 42.,» MM) + a =..P 4. + I; (1') 9‘17: , n 76* ‘7 W use, conservav‘fan 07‘- mass 6?, 3 Q2 ‘I’ 623 M A,v,=ﬁzvz +4314 _ 50 V’ = AZVL+A3V3 = W=/Zﬂﬂé A, /,,,1. 70 esffma/e. R9qie we. Hie Berna/ﬁg 7%, {/W ée/Wm” (dam/(z) 4L. Baas? 4 VI = ﬂange 4— V:— /a 3: 1 lo [sag 2; /z’.ﬂ1](/ N. 5") K§Qx=/{’”2;"")=(””Z§-i g ‘21“) : 40,50011£ time In Now mm: 53.0) we 76’“ t — {2 9(7qu 45?sz 4. (5g)(?79§)ﬁ5§%.5m2](/%:) = m? (40,500 I: Wine) + Fx l»?- CTY‘ ~72 WIN/2. F I K 50 a -.: 72/0WN é— F’h’m} (mm/Waf/ﬁ; Arc-L emkdgy/hﬂ’kahmktWCuja/fv y (WmonyL ml fh //nea>r mmw/wm ermég, ,4, \$.,/ I VlFl/z I42 :1 5 {KEJﬁWEEJﬂSgJﬂzmj 2 67,400N 7‘ = F} 5.37 Water is sprayed radially outward over 180° as indicated + in Fig. P537. The jet sheet is in the horizontal plane. If the jet See‘h'onﬁ ) velocity at the nozzle exit is 20 ft/s, determine the direction and m, Wilma magnitude of the resultant horizontal anchoring force required ' A __ w to hold the nozzle in place i ‘t . FIGURE P5. 37 The conﬁn/ va/umé Ihc [ad'es ﬂ»: may]: and Wa/er bc/wcer; 58675an 0) am! {2) a; xhdicaed #3 7%,; skehaA aéave. AW/luﬁm of 7% y d/iecﬁori campanenf 0W“ #1:. “may momenﬁmv eguaﬁ'on yib/J5 77' z I or 1;” sv/Of(_ié6059)(vz)/1Rd9 = ﬂhﬁgﬁmmsmo) o and F g 0 4,}: :2: Ala/7501190,; of ﬁne X d/Fecﬁ'on Campanern’ of 1%: Una)» momenﬁxm eg act/ion lend: 7‘0 {5 (LIOV. 174/? s at X 77- 1 F = ﬂ/(tgszwﬂvgkmg —.— ﬂaw; (MSG—60:77) 19x ’ a ‘ 2 . I . ﬁ' “"01 F _ (qu/u]5)(0v5in.) (81".) (205)/Z)/ /6 ) Ax , #3 (127\$) (/23) 5/9.; OP AX ~--—— , Campus—ﬂ 5.410 A variable mesh screen produces a lin- ear and axisymmetric velocity proﬁle as indicated in Fig. P5.4.0 in the air ﬂow through a 2~ft-- diameter circular cross section duct. The static: pressures upstream and downstream of the screen are 0.2 and 0.15 psi and are uniformly distributed over the ﬂow cross section area. Neglecting the force exerted by the duct wall on the ﬂowing air, "I = 0‘2 “5i "2 = 0'15 05‘ caicuiate the screen drag force. VI ‘ ‘00 "'5 FIGURE P540 i - -— L Will/IIIIIIIII {Willi/[11117]. Section (1) Section (2) APP/("Cain's of ﬁve axia/ “Myaneﬂf 0/ ﬁle I’Ihea/ mama/um eguaﬁ'm 7‘0 ﬁve f/ow #troujh ﬁve Lon/ml Va/u/ne shown 1;; ﬁre ska r‘o/«i lam/5 7‘0 R --l{(ol{;4[ +ft12€u2 Zﬁ‘rdr : EA ,_ Kx, PA 0 o . ' 7736 value of an,“ may be Oé'lﬂlﬂed {Mm comer-mill”: of may: as {ct/laws ' R 2 M 1:2 =p/(um 27mg” ’7‘ I R 0 77m: 5, = i4 D,“ 12 max 7'“— (Zjigy l’zdr O me / 1 2 2 7. = 40023,? flu moi-f (29} / AS -2170.Wf/_7 /§4f9{2ﬁ L K" ( flair ( 1432/ 5 51 2 _ z _ + 0.2 I}; 7? (2H) 144117").ﬁ/515)1r(28)144£) ‘ 4 H‘ = 3 VI =7 E l00£7=l507tf 2 . 5.453 In a laminar pipe ﬂow that is fully det. average velocity, E, with the axial direction mo— veloped, the axial velocity proﬁle is parabolic, thatis, r2 2 c 1.— —— u “l (M as is illustrated in Fig. P543. Compare the axial direction momentum ﬂowrate calculated with the velocity distribution taken into account. FIGURE P543 7716 axia/ dlfecﬁon meme/714w” [/owmfc based on a um'émq magi}, pmﬁ/e wifh 6: = 12 [5 MI: :1 [l J :: “Z Z X’W’ﬁrm flu/4 ﬂu ’lrl? 7718 axial direcy‘l‘on mower/ﬁrm [/owmfe based on Me non. uniform paraéa/io vc And/y pro ﬁ/c x3 K / 7. MP :[u “Zwrdr : Mlzﬂ‘kz/Z, r-7/r r X170”- (0 ﬂ 4; /—~ ~jd~ a 0 (a R 4) 7— 2 u 7/")? Min“? " W C (071/010?) 3 To own)» a Elwyn/1;); befween a 4440’ ug we use 7% conservm‘ion of mass egaaﬁ'on as 7(0//aw.\$ was/Mym + 0 77w; a z: 91:: Z M}: = 4 [7511’]? g f. M]: . {nm— 37 3 X’um m anhém. 5—43 mentum ﬂowrate calculated with the nonuniform 5.53 The exit plane of a 0.20-m-diameter pipe is partially blocked by a plate with a hﬂlﬁ in it that produces a OJO-m—di- meter strum as shown in Fig. P553. The water velocity in the pipe is 5 m/s. Gravity and viscous effects are negligible. Dcwr- mine the force needed to hold the plate against the pipe. 5 mls IFIGUIRE P5.53 The x*compoﬂem‘ of #he Mame/Wm eql/ah'on For H76 oom'ra/ Volume show” is V, ‘ __ _ .E 9., _I F .3- I_. y V3. [uevhdmrx Mfg Lx E3 cs I._.. __ __ ___'(13 P150 v, p(-V.)ﬂ. napvlnl m4 — F) "3 where F13 {/78 ﬁne in bold {hep/01‘s. 7771/5J F = pﬂ, WWW), -(>Vz"ﬂz =,o,ﬂ, MHZ—V1) (n where 175 = e/ilV, =97? %[g(0:2m31] (5%) z 57% 14150J A,V, =Azlé 0,- V; = (A,/Az.)V, ‘49, /Dz)"\4=(o.2m/0./m)1(5%)=20m,<§ In add/5m, from H73 Bernoulli eqz/ah'm} 70, +-z'—€V,1 = ﬂz+%€\/22 \$0 Md MW] 70;“? Thu-3 from EMU F s [.87x105%[.g(0.2m)1] +157 é?- ( 5% “20%) = 3J5an 5.54 Two water jets of equal size and speed strike ewh other as shown in Fig. P554. Determine the speed, V, and direction, 6, 7 of the resulting combined jet. Gravity is negligible. x I FIGURE P5.54 For Hm Confrbl vokume Shown TM 4hr. Skd’o‘? above “Hm: Urea! moonth equaﬁOA For 11m: x and ydtmc-Hahs are, 'Foy' +he, x dt'recﬁom szszAz +(VCOSGDQVA = O (I) (Md 1"Dv Hat )1 citric.th ~Vle‘A,+(v sinthA : 0 (2) Also “For conseva on 0? M55 we. have PIVIAI “f PM: " PM = 0 (3) From Ecbs. I and 2. we. qe+ Viki = (£0.56 = (0+9 WA. SM 6 1 \$0 -| '1 _, z (0.1” 6 = «3“ V1“: = (30+, (WC-\$37" T , = 45° WA: mﬁ)‘ W (w? L} NOW, wmb‘ning E%S.1 and 3 we gel- -Vz‘A‘a— Vsine (V,A,+\12AL) -.- O of '2. v: M. sin e (V, A, + 01 AL) V 3 min," (33H) _.-——— _.._—-——--— ‘1. (Sinqs‘, ) [00 fit) ’W(O.tﬁ)1’+<10 Tr (OJ-H) ] and ‘f ’+ V : 107 {i :3" 5 54:5 )5. 6% 5.65 A 3-insd-iameter horizontal jet of water stokes a flat plate as inclicated in Fig. P165. Determine the jet velocity if a l()--lb 3 in, ' horizontal force is required to (a) hold the plate stationary, ' F _ 10 lb (b) allow the plate to move at a constant speed of 10 ft/s to the A _ rig-ht. (a) (b) FIGURE 95.65 The comlro/ Volume Sham/n Ih f/ze 565ch I}: (Med. 721\$ sfdﬁ'anary p/a/o can i: cam/Hereo/ ﬁfn‘. A/y/z'caﬁ’on of Me hay/yonfa/ (W X« dlfecﬁbn Com/manenf' of ﬁle 0364,, momel/qu/VM egmﬂm/r y/e/dj "uh/our A] z ’51,): 0r é { M, =(/ 51") 3 FAX I” ’4/ r7 7:42 Thus if L [1 :7. M.“ k V z I __...___..__ {.94 {131: 17,8142: / ’5 - {1‘3 6‘(/2 1}). z 5/” ’9‘ a of /0 2 70 "7 j “‘2 V! (4 = o ' , E. 57(dﬁ0nay lo/ach When Me y/az‘e move; 1‘0 Me r/y/n‘ wifh a speed 1 U= /0 151‘ , ' .17 7%e xa direczé'on comanenf of 71/18 0): ear momem‘um 67414291”, y/e/dj —(uI—U),o(u,—U)/ll = ~63 0/” F 1;: IX L ‘uI—U=_L)=_F;LX) A 2 and IF log. 1 ‘ f #70, a =/___¢__'4X2 1» 'U’ _ f- {7‘ __ {2" I vote) - #12? 7L- /0? - 20.23. ’7‘ mall/h; p/m‘c ...
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## This note was uploaded on 05/04/2010 for the course TAM 335 taught by Professor Whoever during the Spring '08 term at University of Illinois at Urbana–Champaign.

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HW#5%20solution%20corrected - V 15 m/s.— Area = 0.3 m2...

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